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In a Fraunhofer diffraction experiment at a single slit using a light of wavelength $400 \mathrm{~nm}$, the first minimum is formed at an angle of $30^{\circ}$. The direction $\theta$ of the first secondary maximum is given by
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Verified Answer
The correct answer is:
$\sin ^{-1}(1 / 4)$
Given, wavelength of used light,
$$
\lambda=400 \mathrm{~nm}
$$
For minima in Fraunhofer diffraction,
$\therefore \quad d \sin 30^{\circ}=1 \times 400$
$\Rightarrow \quad d=800 \mathrm{~nm}$
For maxima in diffraction,
$$
d \sin \theta=(2 n+1) \frac{\lambda}{2}
$$
For first maxima $n=0$
$$
\begin{array}{ll}
\therefore & d \sin \theta=(0+1) \frac{\lambda}{2} \\
\Rightarrow & d \sin \theta=\frac{\lambda}{2} \\
\Rightarrow & 800 \sin \theta=\frac{400}{2} \\
\Rightarrow & \sin \theta=\frac{1}{4} \\
\Rightarrow & \theta=\sin ^{-1}\left(\frac{1}{4}\right)
\end{array}
$$
$$
\lambda=400 \mathrm{~nm}
$$
For minima in Fraunhofer diffraction,
$\therefore \quad d \sin 30^{\circ}=1 \times 400$
$\Rightarrow \quad d=800 \mathrm{~nm}$
For maxima in diffraction,
$$
d \sin \theta=(2 n+1) \frac{\lambda}{2}
$$
For first maxima $n=0$
$$
\begin{array}{ll}
\therefore & d \sin \theta=(0+1) \frac{\lambda}{2} \\
\Rightarrow & d \sin \theta=\frac{\lambda}{2} \\
\Rightarrow & 800 \sin \theta=\frac{400}{2} \\
\Rightarrow & \sin \theta=\frac{1}{4} \\
\Rightarrow & \theta=\sin ^{-1}\left(\frac{1}{4}\right)
\end{array}
$$
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